Hard Limit Question
- Thread starter peblez
- Start date
Let's say instead of the cube root we have a square root. How would you solve it? You would rationalize the numerator to get a difference of squares and we can cancel a bunch of things by multiplying by "1". Similarly, with this question, try multiplying by "1" so that you'll get a difference of CUBES in the numerator:
\(\displaystyle a^{3} - b^{3} = \left(a - b\right)\left(a^{2} + ab + b^{2}\right)\)
So imagine:
\(\displaystyle a = \sqrt[3]{x^{3} + 5x^{2}}\) and \(\displaystyle b = x\). Now see what you can do. Remember, you're trying to get the numerator to be the difference of cubes.
Edit: Wow this gets messy
\(\displaystyle a^{3} - b^{3} = \left(a - b\right)\left(a^{2} + ab + b^{2}\right)\)
So imagine:
\(\displaystyle a = \sqrt[3]{x^{3} + 5x^{2}}\) and \(\displaystyle b = x\). Now see what you can do. Remember, you're trying to get the numerator to be the difference of cubes.
Edit: Wow this gets messy
Hello, peblez!
Rationalizing with cube roots is a very messy matter.
I'll try to explain . . .
We're expected to know: \(\displaystyle \:a^3\,-\,b^3\:=\a\,-\,b)(a^2\,+\,ab\,+\,b^2)\)
But how does this apply to a problem with cube roots?
\(\displaystyle \text{Replace }a \text{ with }p^{\frac{1}{3}}\;\text{ ... and }b\text{ with }q^{\frac{1}{3}}\)
The right side is: \(\displaystyle \:\left(p^{\frac{1}{3}}\,-\,q^{\frac{1}{3}}\right)\left(p^{\frac{2}{3}}\,+\,p^{\frac{1}{3}}q^{\frac{1}{3}}\,+\,q^{\frac{2}{3}}\right)\;\) . . . which equals: \(\displaystyle \,p\,-\,q\)
\(\displaystyle \text{Our problem has: }\;\sqrt[3]{x^3\,+\,5x^2} \,-\,x \;=\;\overbrace{\left(x^3\,+\,5x^2\right)^{\frac{1}{3}}}^{p^{\frac{1}{3}}}\,-\,\overbrace{\left(x^3\right)^{\frac{1}{3}}}^{q^{\frac{1}{3}}\)
\(\displaystyle \text{We must use: }\;\underbrace{\left(x^3\,+\,5x^2\right)^{\frac{2}{3}}}_{p^{\frac{2}{3}}}\,+\,\underbrace{\left(x^3\,+\,5x^2\right)^{\frac{1}{3}}\left(x^3\right)^{\frac{1}{3}}}_{p^{\frac{1}{3}}q^{\frac{1}{3}}}\,+\,\underbrace{\left(x^3\right)^{\frac{2}{3}}}_{q^{\frac{2}{3}}}\)
So we will multiply top and bottom by: \(\displaystyle \x^3\,+\,5x^2)^{\frac{2}{3}}\,+\,x(x^3\,+\,5x^2)^{\frac{1}{3}}\,+\,x^2\)
. . And the algebra gets even messier . . . Good luck!
Rationalizing with cube roots is a very messy matter.
I'll try to explain . . .
We're expected to know: \(\displaystyle \:a^3\,-\,b^3\:=\
a\,-\,b)(a^2\,+\,ab\,+\,b^2)\)But how does this apply to a problem with cube roots?
\(\displaystyle \text{Replace }a \text{ with }p^{\frac{1}{3}}\;\text{ ... and }b\text{ with }q^{\frac{1}{3}}\)
The right side is: \(\displaystyle \:\left(p^{\frac{1}{3}}\,-\,q^{\frac{1}{3}}\right)\left(p^{\frac{2}{3}}\,+\,p^{\frac{1}{3}}q^{\frac{1}{3}}\,+\,q^{\frac{2}{3}}\right)\;\) . . . which equals: \(\displaystyle \,p\,-\,q\)
\(\displaystyle \text{Our problem has: }\;\sqrt[3]{x^3\,+\,5x^2} \,-\,x \;=\;\overbrace{\left(x^3\,+\,5x^2\right)^{\frac{1}{3}}}^{p^{\frac{1}{3}}}\,-\,\overbrace{\left(x^3\right)^{\frac{1}{3}}}^{q^{\frac{1}{3}}\)
\(\displaystyle \text{We must use: }\;\underbrace{\left(x^3\,+\,5x^2\right)^{\frac{2}{3}}}_{p^{\frac{2}{3}}}\,+\,\underbrace{\left(x^3\,+\,5x^2\right)^{\frac{1}{3}}\left(x^3\right)^{\frac{1}{3}}}_{p^{\frac{1}{3}}q^{\frac{1}{3}}}\,+\,\underbrace{\left(x^3\right)^{\frac{2}{3}}}_{q^{\frac{2}{3}}}\)
So we will multiply top and bottom by: \(\displaystyle \
x^3\,+\,5x^2)^{\frac{2}{3}}\,+\,x(x^3\,+\,5x^2)^{\frac{1}{3}}\,+\,x^2\). . And the algebra gets even messier . . . Good luck!
\(\displaystyle \L\ (x^3 + 5x^2)^{\frac{1}{3}} - x = x[(1 + \frac{5}{x}\)^{\frac{1}{3}} - 1]\)
Now, Soroban, above, explained the following step:
Multiply it by \(\displaystyle \L\ \frac{(1 + \frac{5}{x}\)^{\frac{2}{3}} + (1 + \frac{5}{x}\)^{\frac{1}{3}} + 1}{(1 + \frac{5}{x}\)^{\frac{2}{3}} + (1 + \frac{5}{x}\)^{\frac{1}{3}} + 1}\\)
So \(\displaystyle \L\ x[(1 + \frac{5}{x}\)^{\frac{1}{3}} - 1]\) becomes:
\(\displaystyle \L\ \frac{5}{(1 + \frac{5}{x}\)^{\frac{2}{3}} + (1 + \frac{5}{x}\}^{\frac{1}{3}} + 1}\\)
...by the difference of cubes rule. The limit can now be evaluated.
Now, Soroban, above, explained the following step:
Multiply it by \(\displaystyle \L\ \frac{(1 + \frac{5}{x}\)^{\frac{2}{3}} + (1 + \frac{5}{x}\)^{\frac{1}{3}} + 1}{(1 + \frac{5}{x}\)^{\frac{2}{3}} + (1 + \frac{5}{x}\)^{\frac{1}{3}} + 1}\\)
So \(\displaystyle \L\ x[(1 + \frac{5}{x}\)^{\frac{1}{3}} - 1]\) becomes:
\(\displaystyle \L\ \frac{5}{(1 + \frac{5}{x}\)^{\frac{2}{3}} + (1 + \frac{5}{x}\}^{\frac{1}{3}} + 1}\\)
...by the difference of cubes rule. The limit can now be evaluated.